Question

$H-C\equiv C-H\underset{Excess}{\overset{NaN{H}_2}{\to }}[C\equiv C{]}^{2-}2N{a}^{+}\underset{Excess}{\overset{C{H}_{3}C{H}_{2}Br}{\to }}A$

In the above reaction what is product A?

1. 3-hexyne

2 1-hexyne

Choose the correct option from 1 and 2.

Answer 1

When acetylene is treated with an excess of sodamide, both the acidic hydrogen atoms of acetylene are abstracted.
This is followed by nucleophilic attack on ethyl bromide which is taken in excess.

$H-C\equiv C-H\underset{Excess}{\overset{NaN{H}_2}{\to }}[C\equiv C{]}^{2-}2N{a}^{+}\underset{Excess}{\overset{C{H}_{3}C{H}_{2}Br}{\to }}C{H}_{3}C{H}_{2}C\equiv CC{H}_{2}C{H}_3$

Hence, the correct option is $1$.